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lina2011 [118]
3 years ago
5

HELP ME PLEASE I DON'T UNDERSTAND

Mathematics
1 answer:
hram777 [196]3 years ago
7 0

Answer:

For #9 it can be 4,6, or 7

#10 is angle 3

#11 is 134

#12 is 49

Step-by-step explanation:

#9 just look at the picture given and see which one is opposite from it

#10 look at which one is beside it but that the ’half‘ circles are facing each other

#11 and #12 subtract the angles they give you and see if they are next to each other and if they are that subtract the number they give you by 180-the degrees of a line

You might be interested in
Which of the following sets are subspaces of R3 ?
Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

\to A={(x,y,z)|3x+8y-5z=2} \\\\\to  for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)

                                        =3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)

The set A is not part of the subspace R^3

for point B:

\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon  B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)

                                             =-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0

\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon  B

The set B is part of the subspace R^3

for point C: \to C={(x,y,z)|x

In this, the scalar multiplication can't behold

\to for (-2,-1,2) \varepsilon  C

\to -1(-2,-1,2)= (2,1,-1) ∉ C

this inequality is not hold

The set C is not a part of the subspace R^3

for point D:

\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)

The scalar multiplication s is not to hold

\to for (-4, 1,2)\varepsilon  D\\\\\to  -1(-4,1,2) = (4,-1,-2) ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace R^3

For point E:

\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to  a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\

The  x_1, x_2 is the arbitrary, in which ax_1+bx_2is arbitrary  

\to a(x_1,0,0)+b(x_2,0,0) \varepsilon  E

The set E is the part of the subspace R^3

For point F:

\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon  F \\\\\to  a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))

The x_1, x_2 arbitrary so, they have ax_1+bx_2 as the arbitrary \to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F

The set F is the subspace of R^3

5 0
3 years ago
I have an idea on how to do this I just need to make sure it’s right or not but I’ll mark you!
Akimi4 [234]

Answer:

I think it's 300

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Which function is a quadratic function?
DerKrebs [107]
The correct answer is: [C]:  " c(x) = –8x²<span> + 3x – 5 " .
_________________________________________________________
  The quadratic function takes the form of:
_________________________________________________________
  " f(x) =  ax</span>²  + bx + c " .
_________________________________________________________
Answer choice:  [C]:  " c(x) = –8x² + 3x – 5 "  ;

  is the only answer choice provided that takes that form of the quadratic function:

  " f(x) =  ax²  + bx + c " :

in which:  " a = -8 " ; " b = 3 " ;  " c = -5 " .
____________________________________________________
5 0
2 years ago
Help on this problem?? please?
Reptile [31]

MrBillDoesMath!


Answer to #4:  81/256 * s^8 * t^ 12


Comments:

(7x^3) ^ (1/2)   =  7 ^ (1/2)  *  x^(3/2)   where  ^(1/2) means the square root of a quantity. The answer written (7x^3) is NOT correct.

---------------------

(1)             (27s^7t^11)^ (4/3)  

               = 27^(4/3) * (s^7)^(4/3) * (t^11)^ (4/3)


As 27 = 3^3, 27 ^(4/3) = 3^4 = 81

               

(2)  (-64st^2)^ (4/3)  =     (-64)^(4/3) * (s^4/3) * t(^8/3)


As 64 = (-4)^3,  (-64)^(4/3) = (-4)^4 = +256                                        


So (1)/(2) =

81 * s^(28/3)* t^(44/3)

-------------------------------     =

256 s^(4/3) * t^((8/3)


81/256 *  s ^ (28/3 - 4/3) * t^(44/3 - 8/3) =


81/256 * s^(24/3) * t (36/3) =

81/256 * s^8        * t^ 12



MrB

5 0
3 years ago
The sum of two numbers is 24 and their difference is 2. What are the numbers?
Musya8 [376]
Its 4 and 6
4 X 6= 24 
6-4= 2


8 0
3 years ago
Read 2 more answers
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